音声認識結果が表示されない場合のメッセージの印刷

Question

私は音声認識を使用して色のリストを確認しますが、リストにないと「色が見つかりません」と表示された場合は「色Found '私は各メッセージを一度しか表示したくない。 私が抱えている問題は、「色が見つかりません」というメッセージを正しく表示する方法です。

# speech recognition 

import speech_recognition as speech 

#a lot of variables used for loops and conditions 
voice = speech.Recognizer() 
condition = False 
condition2 = False 
condition3 = False 
boolean = False 
counter = 0 
counter2 = 0 

while condition == False: 
#with microphone input as the input 
with speech.Microphone() as source: 
    print("Say something!") 

    #variable = the input from the microphone 
    audio = voice.listen(source) 
    voiceRecog = voice.recognize_google(audio) 
    try: 
     #print the word that was heard 
     print("\nYou said: " + voiceRecog) 
    except speech.UnknownValueError: 


     #if couldnt recognise then print this msg 
     print("Sorry, we couldn't make that out. Try again!") 

    except speech.RequestError as e: 
     print("Could not request results from Google Speech Recognition service; {0}".format(e)) 


    #list of colours for comparing 
    colours = ["blue", "red", "Green", "Orange", "yellow", "pink", "purple", "black"] 

    #loop to check the word spoken against each word in the list 
    for i in range(len(colours)): 



     #so that if colour found it doesnt claim colour not found 
     if boolean == False: 
      condition2 = False 

     #increase counters by 1 and change booleans to True 
     #if spoken word matches any words in the list then output 
     if colours[i] == voiceRecog: 

      boolean = True 
      condition2 = True 
      condition3 = True 
      counter += 1 
      counter2 += 1 

      print("\nColour Found!\n") 

     #if user says "quit" then all booleans = True (exit all loops) 
     elif voiceRecog == "quit": 

      boolean = True 
      condition = True 
      condition2 = True 
      condition3 = True 

     #if none of other conditions check value of i and of con2 
     else: 
      #if end of list is reached and con2 = False then con3 = False 
      if (i + 1) == len(colours) and condition2 == False: 
       condition3 = False 
       print(end = "") 
    #if con3 = False then counter increase and print 'colour not found' 
    if condition3 == False:    
     print("\nColour not found!\n\n") 
     counter += 1 

#once loop exited by saying quit, print attempts and successful attempts 
print("\nYou checked for", counter, "colours and had", counter2, "successful attempts") 

上記は私のコードであり、以下は1つのシナリオです。

何かを言います！

あなたは言った：青

色が見つかりました！

何か言ってください！

あなたは言った：緑

色が見つかりました！

何か言ってください！

あなたは言った：デイブデイブ

は、何かを言います！

あなたは言った：あなたは2色をチェックし、3回の試行が、2つの成功した試みがあるはずです2つの成功した試み

を持っていた

を終了します。

しかし、私はそれを他の方法で回避を行う場合：

は、何かを言います！

あなたは言った：デイブデイブ

カラーが見つかりません！

何か言ってください！

あなたは言った：スティーブスティーブ

色が見つかりません！

何か言ってください！

あなたは言った：赤

色が見つかりました！

何か言ってください！

あなたは言った：黒

色が見つかりました！

何か言ってください！

あなたは言った：あなたは4色をチェックしていた

を辞め2つの成功した試み

私はそれがこれをやっている理由を知っているが、私はそれを回避する方法を把握することはできません。

Answer 1

こんにちは、私はPythonのsyntaxes..hereについて知らないC＃のプログラマがC＃で、私の考えですよ、それが `

variable succes_Counter 
variable unsuccess_Counter = 0 
boolean color_not found = true 
for i in range(len(colours)): 
    { 

      if colours[i] == voiceRecog 
      { 
       print("color found + colours[i] + ") 
       succes_Counter += 1 

       //HERE IT EXISTS FROM THIS LOOP 

       break 
      } 
      elseif voiceRecog == "quit" 

      //HERE IT EXISTS FROM THIS LOOP 
       break 


      else 
       if color_not found == true 
       { 
       print(" color not found") 

       unsuccess_Counter +=1 

       color_not found = false 
       } 
     } 


      color_not found = true 
      print("\nYou checked for", unsuccess_Counter, "colours and had", succes_Counter, "successful attempts") 

Answer 2

# speech recognition 

import speech_recognition as speech 

#a lot of variables used for loops and conditions 
voice = speech.Recognizer() 
# Maintain status of each attempt. You can just use two variables successful_attempts and total_attempts also but status list might help you in maintaining more things like a list of tuples (color, result) etc 
status = [] 
# successful_attempts, total_attempts = 0, 0 

while True: 
    #with microphone input as the input 
    with speech.Microphone() as source: 
     print("Say something!") 

     #variable = the input from the microphone 
     audio = voice.listen(source) 
     voiceRecog = voice.recognize_google(audio) 
     try: 
      #print the word that was heard 
      print("\nYou said: " + voiceRecog) 
     except speech.UnknownValueError: 


      #if couldnt recognise then print this msg 
      print("Sorry, we couldn't make that out. Try again!") 

     except speech.RequestError as e: 
      print("Could not request results from Google Speech Recognition service; {0}".format(e)) 


     #list of colours for comparing 
     colours = ["blue", "red", "Green", "Orange", "yellow", "pink", "purple", "black"] 

     if "quit" in voiceRecog: 
      break 
     elif voiceRecog in colours: 
      status.append(1) 
      # successful_attempts += 1 
      print("\nColour Found!\n") 
     else: 
      status.append(0) 
      print("\nColour not found!\n\n") 

     # total_attempts += 1 

#once loop exited by saying quit, print attempts and successful attempts 
print("\nYou checked for", len(status), "colours and had", sum(status), "successful attempts") 
#print("\nYou checked for", total_attempts, "colours and had", successful_attempts, "successful attempts") 

を作品にこれを試してください